Ellipse and Tangentials

Let me propose a problem.

Q. When the product of the slopes of 2 tangential on the ellipse $\large\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$ from dot $P(p,q)$ outside of the curve is $\alpha,$ equate $α$ using $a, b, p, \quad \text{and} \quad q.$

|Solution|

We can think of the circle $x^2+y^2=a^2$ , made by multiplying $\frac{a}{b}$ on $y$ -coordinate.

Multiply $\frac{a}{b}$ on the $y$ -coordinate of dot $P$ , and define a new dot $P'(p,\frac{a}{b} q).$

The tangential on the circle $x^2+y^2=a^2$ from dot $P’$ is $y=m(x-p)+\frac{a}{b} q.$

The distance from zero point to the dot of contact $d$ is equal to $a$ , so $d=\frac{|-mp+\frac{a}{b} q|}{\sqrt{m^2+1}}=a.$

Simplified, $a^2 m^2+a^2=m^2 p^2-\frac{2ma}{b} pq+\frac{a^2}{b^2} q^2, \quad m^2 (p^2-a^2 )-\frac{2ma}{b} pq+(\frac{a}{b})^2 (q^2-b^2 )=0.$

This is the quadratic equation on $m$ , and the product of every possible $m$ is $\large(\frac{a}{b})^2 (\frac{q^2-b^2}{p^2-a^2}).$

Since we’d multiplied $\frac{a}{b}$ on the $y$ -coordinate, product of the 2 slopes of the tangential on the circle $x^2+y^2=a^2$ from dot $P’$ is $(\frac{a}{b})^2 α.$

Thus, $α=\large\frac{q^2-b^2}{p^2-a^2}.$

|Application|

Since $αp^2-αa^2=q^2-b^2, αp^2-q^2=αa^2-b^2,$ trace of dot $P(p,q)$ is the quadratic curves.

We need only to observe when $\alpha=-\dfrac{1}{2}$ .

Since $p^2-a^2=2b^2-2q^2, p^2+2q^2=a^2+2b^2,$ the trace of the dot $P$ is an ellipse of which the center is the dot $O(0,0)$ .

In this case, $a^2+2b^2=6,$ and thus the dot $P$ moves along the ellipse $\dfrac{x^2}{6}+\dfrac{y^2}{3}=1$ . Therefore, the area enclosed by $P$ 's trace is $\sqrt{18}\pi$ , and the value of $k$ is $\boxed{18}$ .

P.S. Also try my another problem 'Ellipse and the Right Angle'.

Relevant wiki: Ellipse

Best solution will be the way we find the locus of director circle.

Here we go.

We know that equation of tangent an ellipse is $y=mx+ \sqrt{a^2m^2+b^2}$

Putting the given values of a and b, we have

$y=mx+\sqrt{4m^2+1}$

$\implies y-mx = \sqrt{4m^2+1}$

$\implies (y-mx)^2 = 4m^2+1$

$\implies y^2+m^2x^2-2mxy = 4m^2+1$

$\implies m^2(x^2-4)-2mxy+(y^2-1)=0$

So we can write $m_{1}m_{2} = \dfrac{-1}{2}$

$\implies \dfrac{y^2-1}{x^2-4} = \dfrac{-1}{2}$

$\implies 2y^2+x^2=6$

Now lenght of semi major axis for this ellipse is = $\sqrt{\dfrac{3}{2}}$

Lenght of semi minor axis = $\sqrt{\dfrac{3}{4}}$

$\implies A= \pi ab$

$\implies A= \pi \sqrt{18}$

$\implies k=\sqrt{18}$